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Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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$\triangle OPA \cong \triangle OSA$
$\Rightarrow \angle 1 = \angle 2$
Similarly, $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^{\circ}$
$\Rightarrow (\angle 1 + \angle 8) + (\angle 4 + \angle 5) = 180^{\circ}$
$\Rightarrow \angle AOD + \angle BOC = 180^{\circ}$_a_1.png)
$\Rightarrow \angle 1 = \angle 2$
Similarly, $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$
Now, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^{\circ}$
$\Rightarrow (\angle 1 + \angle 8) + (\angle 4 + \angle 5) = 180^{\circ}$
$\Rightarrow \angle AOD + \angle BOC = 180^{\circ}$